Question

Question 1 (13 marks) A charge Q is located on the top-left corner of a square. Charges of 2Q, 3Q and 4Q are located on the other corners as shown above in the Figure. Assuming each side of the square is 5cm in length, and the charge Q= 2.5 µC, determine the net force exerted on the 2Q charge due to the other charges.

Answer 1

Answer:

F = 45 (2.4142 i ^ + 4.414 j ^)

  F = 226.40 N,           θ= 61.3

Explanation:

For this exercise we will use that the forces are vectors and we will add them, the force due to being electric charges must comply, go Coulomb's law

         F₁₂ = k q₁ q₂ / r₁₂²

To apply this equation to our case, they indicate that all charges are of the same sign and their value, also the charge Q is located in the upper left corner, unfortunately the diagram of the other charges is not loaded, but the most general is that is in sequence, see attached, for the sum of the forces let's add its components

x-axis (horizontal)

         Fₓ = F₂₁ + F₂₄ₓ

Let's use trigonometry for the force component, as the weathering figure is a square the angle between the force and the x axis is 45

        cos 45 = F₂₄ₓ / F₂₄

         F₂₄ₓ = F₂₄ cos 45

y-axis (vertical)

        $F_{y}$ = F₂₃ + F_{24y}

        sin 45 = F_{24y} / F₂₄

        F_{24y} = F₂₄ sin 45

let's search every distance

The side of the square is worth l = 5cm = 0.05 m

we can find the diagonal with the Pythagorean theorem

         d = √(l² + l²) = l √2

now we can search every force

         F₂₁ = k q₁ q₂ / r₁₂²

         F₂₁ = k Q 2Q / l²

         F₂₁ = k 2Q² / l²

this force points in the positive x direction

         F₂₃ = k q₂ q₃ / r₂₃²

         F₂₃ = k 2Q 3Q / l²

this force points in the direction of the positive y

         F₂₄ = k q₄q₂ / d²

         F₂₄ = k 4Q 2Q / 2 l²

let's find the resultant in each x

X axis

        Fₓ = k 2Q² / l² + k 8Q² / 2 l²    cos 45

        Fₓ = k 2Q² / l² (1 + 4/2 cos 45)

        Fₓ = k 2Q²/ l²   2.4142 i ^

Axis y

       F_{y} = k 6Q² / l² + k 8Q² / 2 l²    sin 45

       F_{y} = k 2Q² / l² (3 + 4/2 sin45)

       F_{y} = k 2Q² / l² (4,414)

       

the resultant force

       F = Fₓ i ^ + F_{y} j ^

       F = k 2Q² / l²      ( 2.4142 i ^ + 4.414 j ^)

       

let's substitute the values

       F = 9 10⁹ 2 (2.5 10⁻⁶) 2 / 0.05²      (2.4142 i ^ + 4.414 j ^)

       F = 45 (2.4142 i ^ + 4.414 j ^)

The result we can give is this form and in the form of module and angle

             

let's use the Pythagorean theorem to find the modulus

            F =√ (Fₓ² +  $F_{y}$²)

            F = 45 √ (2.4142² + 4.414²)

            F = 226.40 N

we use trigonometry for the angle, measured from the x-axis

           tan θθ = Fy / Fx

          θ  = tan⁻¹ (4.414 / 2.4142)

           θ= 61.3